Ta có: \(n_{Ba\left(OH\right)_2}=\dfrac{34,2}{171}=0,2\left(mol\right)\)

a. PTHH: \(Ba\left(OH\right)_2+Na_2SO_4--->BaSO_4\downarrow+2NaOH\)

b. Theo PT: \(n_{BaSO_4}=n_{Ba\left(OH\right)_2}=0,2\left(mol\right)\)

=> \(m_{BaSO_4}=0,2.233=46,6\left(g\right)\)

c. Theo PT: \(n_{Na_2SO_4}=n_{Ba\left(OH\right)_2}=0,2\left(mol\right)\)

=> \(m_{Na_2SO_4}=0,2.142=28,4\left(g\right)\)

=> \(C_{\%_{Na_2SO_4}}=\dfrac{28,4}{240}.100\%=11,83\%\)

PƯ: Ba(OH)₂ + Na₂SO₄ = BaSO₄ + 2NaOH (1)
0,08 -----> 0,08 (mol)
Ta có: n_Ba(OH)2= \(\dfrac{40.34.2\%}{100\%.171}\)= 0,08 (mol)
Theo pư (1) có : n_Na2SO4 0,08 (mol)
=> m_Na2SO4 = 0.08.142 = 11,36 (g)
=> m_ddNa2SO4= \(\dfrac{11,36}{14,2}.100\%\)= 80 (g)

\(n_{Na_2SO_4}=\dfrac{142.10}{100.142}=0,1(mol)\\ Na_2SO_4+Ba(OH)_2\to BaSO_4\downarrow+2NaOH\\ \Rightarrow n_{BaSO_4}=n_{Ba(OH)_2}=0,1(mol);n_{NaOH}=0,2(mol)\\ a,m_{BaSO_4}=0,1.233=23,3(g)\\ b,m_{dd_{Ba(OH)_2}}=\dfrac{0,1.171}{15\%}=114(g)\\ c,C\%_{NaOH}=\dfrac{0,2.40}{142+114-23,3}.100\%=3,44\%\)

Ta có: \(n_{Na_2SO_4}=\dfrac{\dfrac{10\%.142}{100\%}}{142}=0,1\left(mol\right)\)

\(PTHH:Na_2SO_4+Ba\left(OH\right)_2--->BaSO_4\downarrow+2NaOH\)

a. Theo PT: \(n_{BaSO_4}=n_{Ba\left(OH\right)_2}=n_{Na_2SO_4}=0,1\left(mol\right)\)

\(\Rightarrow m_{BaSO_4}=0,1.233=23,3\left(g\right)\)

b. Ta có: \(m_{Ba\left(OH\right)_2}=0,1.171=17,1\left(g\right)\)

Mà: \(C_{\%_{Ba\left(OH\right)_2}}=\dfrac{17,1}{m_{dd_{Ba\left(OH\right)_2}}}.100\%=15\%\)

\(\Leftrightarrow m_{dd_{Ba\left(OH\right)_2}}=114\left(g\right)\)

c. Ta có: \(m_{dd_{NaOH}}=114+14,2-23,3=104,9\left(g\right)\)

Theo PT: \(n_{NaOH}=2.n_{Ba\left(OH\right)_2}=2.0,1=0,2\left(mol\right)\)

\(\Rightarrow m_{NaOH}=0,2.40=8\left(g\right)\)

\(\Rightarrow C_{\%_{NaOH}}=\dfrac{8}{104,9}.100\%=7,63\%\)

a) \(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)

b) \(n_{CuSO_4}=0,4.1=0,4\left(mol\right)\)

\(n_{NaOH}=\dfrac{40}{40}=1\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,4}{1}< \dfrac{1}{2}\) => CuSO₄ hết, NaOH dư

PTHH: CuSO₄ + 2NaOH --> Cu(OH)₂ + Na₂SO₄

_______0,4----->0,8---------->0,4-------->0,4

Cu(OH)₂ --t^o--> CuO + H₂O

0,4------------->0,4

=> m_CuO = 0,4.80 = 32 (g)

c) \(\left\{{}\begin{matrix}m_{NaOH\left(dư\right)}=\left(1-0,8\right).40=8\left(g\right)\\m_{Na_2SO_4}=0,4.142=56,8\left(g\right)\end{matrix}\right.\)

\(a,n_{Na_2SO_4}=0,2\cdot0,2=0,04\left(mol\right);n_{Ba\left(OH\right)_2}=0,2\cdot0,1=0,02\left(mol\right)\\ PTHH:Na_2SO_4+Ba\left(OH\right)_2\rightarrow2NaOH+BaSO_4\downarrow\\ TL:....1.....1......2......1\left(mol\right)\\ BR:.......0,02.....0,02......0,04......0,02\left(mol\right)\)

Vì \(\dfrac{n_{Na_2SO_4}}{1}>\dfrac{n_{Ba\left(OH\right)_2}}{1}\) nên \(Na_2SO_4\) dư, \(Ba\left(OH\right)_2\) hết

\(b,C_{M_{NaOH}}=\dfrac{0,04}{0,2+0,2}=0,1M\)

\(a,Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ b,n_{H_2}=\dfrac{7,437}{24,79}=0,3mol\\ n_{Zn}=n_{H_2SO_4}=n_{ZnSO_4}=n_{H_2}=0,3mol\\ m_{Zn}=0,3.65=19,5g\\ c,C_{M_{H_2SO_4}}=\dfrac{0,3}{0,05}=6M\\ d,m_{ZnCl_2}=0,3.136=40,8g\)

\(a)2Al + 6CH_3COOH \to 2(CH_3COO)_3Al + 3H_2\\ b)n_{Al} = \dfrac{2,7}{27} = 0,1(mol) ; n_{CH_3COOH} = \dfrac{200.10\%}{60} = \dfrac{1}{3}(mol)\\ n_{CH_3COOH} = \dfrac{1}{3}> 3n_{Al} = 0,3 \to CH_3COOH\ dư\\ n_{H_2} = \dfrac{3}{2}n_{Al} = 0,15(mol) \Rightarrow V_{H_2} = 0,15.22,4 = 3,36(lít)\\ n_{CH_3COOH\ pư} = 3n_{Al} =0,3(mol) \Rightarrow m_{CH_3COOH\ pư} = 0,3.60 = 18(gam)\\ c) m_{dd} = 2,7 + 200 - 0,15.2 = 202,4(gam)\\ n_{(CH_3COO)_3Al} = n_{Al} = 0,1(mol)\\ m_{CH_3COOH\ dư} = 200.10\% - 18 = 2(gam)\\ C\%_{(CH_3COO)_3Al} = \dfrac{0,1.204}{202,4}.100\% = `10,08\%\\ \)

\(C\%_{CH_3COOH} = \dfrac{2}{202,4}.100\% = 0,988\%\)